Consider the integral $I=\underset{0}{\overset{10}{\int }}\frac{\left[x\right]e^{\left[x\right]}}{e^{x-1}}dx,$ where $\left[x\right]$ denotes the greatest integer less than or equal to x. Then the value of $I$ is equal to 

$\begin{array}{rcl}\mathrm{I}& =& {\int }_0^{10} \frac{\left[\mathrm{x}\right] {\mathrm{e}}^{\left[\mathrm{x}\right]}}{{\mathrm{e}}^{\mathrm{x}-1}} \mathrm{dx}\\ & =& \mathrm{e}{\int }_0^{10} \frac{\left[\mathrm{x}\right] {\mathrm{e}}^{\left[\mathrm{x}\right]}}{{\mathrm{e}}^{\mathrm{x}}} \mathrm{dx}\\ & =& \mathrm{e}\left[{\int }_0^1 \frac{0 {\mathrm{e}}^0}{{\mathrm{e}}^{\mathrm{x}}} \mathrm{dx}+{\int }_1^2 \frac{1.{\mathrm{e}}^1}{{\mathrm{e}}^{\mathrm{x}}} \mathrm{dx}+{\int }_2^3\frac{2.{\mathrm{e}}^2}{{\mathrm{e}}^{\mathrm{x}}} \mathrm{dx}+...+{\int }_9^{10}\frac{9.{\mathrm{e}}^9}{{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}\right]\end{array}$

$\begin{array}{rcl} & =& \mathrm{e}\left[0+\mathrm{e}{\left(-{\mathrm{e}}^{-\mathrm{x}}\right)}_1^2+2{\mathrm{e}}^2{\left(-{\mathrm{e}}^{-\mathrm{x}}\right)}_2^3+....+9 {\mathrm{e}}^9{\left(-{\mathrm{e}}^{-\mathrm{x}}\right)}_9^{10}\right]\\ & =& \mathrm{e}\left[0+\mathrm{e}\left({\mathrm{e}}^{-1} -{\mathrm{e}}^{-2}\right)+2{\mathrm{e}}^2\left[{\mathrm{e}}^{-2}-{\mathrm{e}}^{-3}\right]+...+9{\mathrm{e}}^9\left[{\mathrm{e}}^{-9}-{\mathrm{e}}^{-10}\right]\right]\\ & =& \mathrm{e}\left[1-{\mathrm{e}}^{-1}+2-2{\mathrm{e}}^{-1}+...+9-9{\mathrm{e}}^{-1}\right]\\ & =& \mathrm{e}\left[\left(1+2+...\left(9\right)\right)-{\mathrm{e}}^{-1}\left(1+2+...+9\right)\right]\\ & =& \mathrm{e}\left[45-\frac{45}{\mathrm{e}}\right]=45\mathrm{e}-45\\ & =& 45(\mathrm{e}-1)\end{array}$